Q. In a series combination $R = 300\ \Omega,\ L = 0.9H,\ C = 2.0\ \mu F,\ \omega = 1000\ {rads}^{- 1}$, the impedance of the circuit is

A) 1300 $\Omega$
B) 900 $\Omega$
C) 500 $\Omega$
D) 400 $\Omega$
Explanation

(c) In series R – L – C circuit, the impedance of the circuit is given by
$Z = \sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}}$
Also, $X_{L} = \omega L,\ \ X_{C} = \frac{1}{\omega C}$
$\therefore\ \ \ \ \ \ \ \ \ \ Z = \sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}$
Given, $\ R = 300\Omega,\omega = 1000\ $rad$s^{- 1},\ L = 0.9\ H,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ C = 20\mu F = 2 \times 10^{- 6}F$
Hence, $Z = \sqrt{(300)^{2}} + \left( 1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{- 6}} \right)^{2}$
$= \sqrt{90000 + (900 - 500)^{2}}$
$= \sqrt{90000 + 160000}$
$= \sqrt{250000} = 500\ \Omega$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.